Question

Experiment III A sample space consists of five simple events with \(P\left(E_{1}\right)=P\left(E_{2}\right)=.15, P\left(E_{3}\right)=.4,\) and \(P\left(E_{4}\right)=2 P\left(E_{5}\right) .\) Find the probability of event \(A=\left\\{E_{1}, E_{3}, E_{4}\right\\}\).

Short answer

Answer: The probability of event A is 0.75.

Step-by-step solution

Find the probability of \(E_{5}\)

Since the sum of probabilities of all simple events in the sample space equals 1, we can write the following equation: \(P(E_1) + P(E_2) + P(E_3) + P(E_4) + P(E_5) = 1\). We are given that: \(P\left(E_{1}\right)=P\left(E_{2}\right)=.15, P\left(E_{3}\right)=.4,\) and \(P\left(E_{4}\right)=2 P\left(E_{5}\right)\). Now, substitute these values into the equation and solve for \(P(E_5)\): \(0.15 + 0.15 + 0.4 + 2P(E_5) + P(E_5) = 1\).

Determine the probability of \(E_{4}\)

Now, solve the equation from Step 1 for \(P(E_5)\): \(2P(E_5) + P(E_5) = 1 - 0.15 - 0.15 - 0.4\) \(3P(E_5) = 0.3\) \(P(E_5) = 0.1\). Since we are given that \(P(E_4) = 2P(E_5)\), we can find the probability of \(E_4\) by substituting the value of \(P(E_5)\): \(P(E_4) = 2 \times 0.1 = 0.2\).

Calculate the probability of event A

Now we have the probabilities for each individual event in event A: \(P(E_1), P(E_3),\) and \(P(E_4)\). To find the probability of event A, we simply add up the probabilities of each individual event: \(P(A) = P(E_1) + P(E_3) + P(E_4)\) \(P(A) = 0.15 + 0.4 + 0.2\) \(P(A) = 0.75\). Therefore, the probability of event A is 0.75.

Key concepts

Sample Space

Understanding the concept of 'sample space' is fundamental in learning about probability. Imagine you're tossing a coin; there are two possible outcomes - heads or tails. These outcomes form what we call the sample space of the experiment, sometimes denoted as 'S'. It is a collection of all possible outcomes.

In our example from the exercise, the sample space consists of five simple events. The term 'simple' implies that these events cannot be broken down into further outcomes. They are often referred to as 'atomic' outcomes because each represents a unique, indivisible possibility within the sample space. This concept is vital because the sum of the probabilities of all simple events within a sample space always equals one. This makes sense because the sample space covers all possible outcomes, meaning something within this set must occur, therefore the total probability is 100%.

Simple Events

When we talk about 'simple events' in probability theory, we're referring to events that cannot be decomposed into smaller, constituent parts. They are the single outcomes that make up the sample space. A simple event is essentially one possibility from the set of all outcomes.

For instance, if we roll a six-sided die, each number that can land face-up (1 through 6) is a simple event. In the problem at hand, each of the five simple events (\(E_{1}\to E_{5}\)) represents a unique possibility in the experiment's sample space. The key here is to recognize that the probability of each simple event is a quantifiable expression of how likely it is to occur; and this can range from 0 (impossible event) to 1 (certain event). This idea forms the backbone of calculating more complex probabilities, by combining the chances of these simple events.

Probability Theory

Probability theory deals with analyzing how likely events are to occur, expressed numerically between 0 and 1. It's the branch of mathematics that governs the behavior of random phenomena.

To make this comprehension as straightforward as possible, let's use the exercise example. Given multiple simple events with known probabilities, we were able to determine the likelihood of a compound event, event A, which consisted of the simple events \(E_{1}, E_{3}, E_{4}\). In probability theory, this addition rule is applied because event A is made up of disjoint events - events that cannot occur simultaneously - making the calculation of its probability a simple sum of the probabilities of its constituent simple events.

Importance of Unity in Probability

The solution demonstrates a vital principle in probability theory: every possible outcome within the sample space collectively holds the certainty of occurrence, hence the total probability summing up to unity (1). This principle enables us to evaluate missing probabilities, as seen with event \(E_{5}\), further enriching our understanding of how likely various outcomes are when faced with uncertainty.